# Triangulating Squares

A square cannot be subdivided into an odd number of triangles, all of the same area.

The problem has appeared in a 1979 issue of the popular Russian magazine *Kvant*, was included into a 1999 selection of articles from Kvant, and also in an exciting collection of solved problem by B. Bollobás.

An obvious tessellation of a square into equal rectangular strips which are then subdivided by one (or both) of the diagonals shows that it is possible to divide a square into any even number of triangles of the same area. We need to prove that, for no odd number, such division is possible. The proof quite unexpectedly makes use of the 2-adic norm.

Let a triangle in the Cartesian plane be defined by vertices _{1}, y_{1}),_{2}, y_{2})._{1}y_{2} - x_{2}y_{1}|._{2} ≤ 2.

The proof is very accessible, but skirts one important fact which is beyond elementary: algebraically, **R**⊂**Q**_{p}, for any prime p, 2 in particular. This is even though the absolute values |.| and |.|_{p} induce essentially different topologies on the set **R** of reals.

We mark all the points (x, y) of the plane in one of three ways:

- points with |x|
_{2}< 1 and |y|_{2}< 1 are marked 0, - points with |x|
_{2}≥ 1 and |x|_{2}≥ |y|_{2}are marked 1, - points with |y|
_{2}≥ 1 and |y|_{2}> |x|_{2}are marked 2.

We write m(x, y) for the markings of point (x, y). Observe that **R**^{2} receives one of the three markings. This coloring of the plane has two important properties:

If m(a, b) = 0, then m(a + x, b + y) = m(x, y). The translation by a vector (point) of markings 0 does not change the marking of a point.

Indeed, by the strong triangle inequality, |a + x|

_{2}= max{|a|_{2},|x| meaning that either_{2}} = max{1, |x|_{2}},|a + x| or |x|_{2}= |x|_{2},_{2}< 1 (or both of course.) Similarly,|b + y| or |y|_{2}= |y|_{2},_{2}< 1. Thusm(x, y) = 0 impliesm(a + x, b + y) = 0. m(x, y) = 1 impliesm(a + x, b + y) = 1, and the same holds for the points marked 2.- No three points with distinct markings are collinear. This follows from the fact that the are of a triangle with vertices of all three markings could not be 0. In fact, the area of such a triangle is at least 2!
Indeed let there be a triangle with all three markings. (We call such triangles complete.) Translate it into triangle T with the 0 vertex at the origin. The translation changes neither area not the markings of the vertices. Let the vertices of T be (0, 0),

(a, b) and(u, v). Without loss of generality, we may assume that(a, b) is marked 1 and(u, v) is marked 2. In particular,|a| and_{2}≥ |b|_{2}≥ 1|v| It follows that_{2}> |u|_{2}≥ 1.|av| And of course_{2}= |a|_{2}|v|_{2}> |b|_{2}|u|_{2}= |bu|_{2}.|av| We now have_{2}≥ 1.|Area(T)|

_{2}= |1/2 (uv - bu)|_{2}= 2|av - bu|_{2}= 2|av|_{2}≥ 2,as claimed. All that remains to be shown is that any triangulation of a square into any number of triangles contains a complete triangle.

For convenience, consider a unit square with vertices O(0, 0), A(0, 1), B(1, 0), and C(1, 1). Clearly,

m(O) = 0, m(A) = 1 whilem(B) = 2. Since no three differently marked points are collinear. OA may only contain points marked 0 or 1.

Now, we apply the Index Lemma. Let's count the number of the triangle sides marked 01 on the boundary of the unit square OACB, similarly to what has been done in the proof of Sperner's Lemma. The edge OA contains an odd number of such sides. The edges AC, BC (0, in fact), and OB (also just 0) contain an even number of sides 01, making the the total number of the 01 edges on the boundary of the square odd. Therefore, by the Index Lemma, the number of complete triangles in the triangulation is also odd. In particular, there is at least one complete triangle, and we are done.

### References

- B. Bollobás,
*The Art of Mathematics: Coffee Time in Memphis*, Cambridge University Press, 2006, #108 - B. Bekker, S. Vostokov, Yu. Ionin,
__2-adic numbers__, in*Kvant Selecta: Algebra and Analysis I*(S. Tabachnikov ed.), AMS, 1999, 99-110

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